Originally Posted by

**georgel** First of all, thank you for your responses.

This is what I did:

$\displaystyle u(x, y)=X(x)Y(y)$

$\displaystyle X''Y + Y''X=0$

$\displaystyle \frac{X''}{X}=-\frac{Y''}{Y}=-\lambda^2$, $\displaystyle \lambda \geq 0$

According to my textbook, we use $\displaystyle \lambda$ _squared_ because it's easier to work with it later, and the minus is according to the theorem of non-negativity of eigenvalues (I'm not sure that's the correct english term.) so basically, I'm checking $\displaystyle \lambda =0$ and $\displaystyle \lambda > 0$.

$\displaystyle \lambda=0$ gives me:

$\displaystyle X''=0$, that is, $\displaystyle X=Ax+B$.

$\displaystyle X(0)=0$ gives me $\displaystyle B=0$.

So, I have $\displaystyle X=Ax$, which is not a trivial solution. I get the same when I do it like you said, $\displaystyle X''+kX=0$.

What did I do wrong?

$\displaystyle \lambda>0$:

Characteristic equation:

$\displaystyle x^2=-\lambda^2$

$\displaystyle x=+- \sqrt{-\lambda^2}=+-i\lambda$

$\displaystyle X=Csin(\lambda x)+Dcos(\lambda x)$

$\displaystyle X(0)=0$ gives me $\displaystyle D=0$. So, $\displaystyle X=Csin(\lambda x)$.

I'm really confused, please help.